Question

A parallel-plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and then the battery is disconnected. Now, the slab is pulled out slowly at $t=0$. If at time $t$, the capacitance of the capacitor is $C$ and potential difference between the plates of the capacitor is $V$, then which of the following graphs is/are correct.

• A.
• B.
• C.
• D.

Answer 1

Answer: (A), (C)

We need two relations one is between $C$ and $t$ and other is between $V$ and $t$.

Let $C$ be the capacitance with dielectric fully inserted.

Now the scenario is as shown

$C_{eq}=\frac{K{\epsilon }_0(A-vtb)}{d}+\frac{{\epsilon }_{0}vtb}{d}$
$\begin{array}{cc}{C}_{eq}& =\frac{K{\epsilon }_{0}A}{d}-\frac{K{\epsilon }_{0}vtb}{d}+\frac{{\epsilon }_{0}vtb}{d}\\ & =-t[\frac{K{\epsilon }_{0}vb}{d}-\frac{{\epsilon }_{0}vb}{d}]+\frac{K{\epsilon }_{0}A}{d}\\ & =-t\left[\frac{{\epsilon }_{0}vb}{a}\right[K-1\left]\right]+\frac{K{\epsilon }_{0}A}{d}\end{array}$
as the equation we obtained is in the form of $y=-mx+c$, option a is correct.

Again we have,
$Q=CV$
$C=\frac{Q}{V}$
$C\alpha \frac{1}{V}$
Therefore option C is also correct.